Define the concepts of a convergent series and an absolutely convergent series. For which $a ∈ \mathbb R$ number
$$\sum_{n=1}^{\infty}{\frac{(-1)^{n}}{n^a}}$$
on is convergent, and for which it converges absolutely?
I want to use d'Alebert for this. And check this for periods for the $a$.
1) $a\in (0,1)$
2) $a\in (1,\infty)$
3) $a=0$
4) $a\in (-1,0)$
5) $a\in (-\infty,-1)$
I stared calculated $lim_{x\to \infty}{{\frac{a_n+1}{a_n}}}$
$lim_(x\to \infty){\frac{\frac{(-1)^{n+1}}{(n+1)^a}}{\frac{(-1)^{n}}{n^a}}}=lim_(x\to \infty){\frac{(-1)^{n+1}}{(n+1)^a}}{\frac{n^a}{(-1)^{n}}}=-lim_(x\to \infty){\frac{n^a}{(n+1)^a}}$
This is for a convergent series. Equation for a absolute series will be without $-$. This place where I'm stuck. What is next step?
Recall that
$$\sum_{n=1}^{\infty}{\left|\frac{(-1)^{n}}{n^a}\right|}=\sum_{n=1}^{\infty}{\frac{1}{n^a}}$$
converges, by integral test, if and only if $a>1$ and
$$\sum_{n=1}^{\infty}{\frac{(-1)^{n}}{n^a}}$$
converges by alternating series test for $a>0$ and diverges otherwise since $a_n \not \to 0$.